Problem: Let $f(x)=\begin{cases} \ln(-x)+3&\text{for }x < -3 \\\\ \ln(-x+3)&\text{for }-3 \leq x < 3 \end{cases}$ Is $f$ continuous at $x=-3$ ? Choose 1 answer: Choose 1 answer: (Choice A) A Yes (Choice B) B No
For $f$ to be continuous at $x=-3$, we need $\lim_{x\to -3}f(x)$ and $f(-3)$ to exist and be equal. Since $-3\geq -3$, the rule that applies to $x=-3$ is $\ln(-x+3)$. So $f(-3)=\ln(-(-3)+3)=\ln(6)$. Now let's analyze $\lim_{x\to -3}f(x)$. Finding $\lim_{x\to -3^{ +}}f(x)$ For $x$ -values larger than $-3$, the appropriate rule for $f(x)$ is $\ln(-x+3)$. Since $\ln(-x+3)$ is continuous for $-3\leq x<3$, any limit in this interval is equal to the expression's value at that limit: $\begin{aligned} &\phantom{=}\lim_{x\to -3^{ +}}f(x) \\\\ &=\lim_{x\to -3^{ +}}[\ln(-x+3)] \gray{\ln(-x+3)\text{ is the rule for }x>-3} \\\\ &=\ln(-(-3)+3) \gray{\ln(-x+3)\text{ is continuous at }x=-3} \\\\ &=\ln(6) \end{aligned}$ Finding $\lim_{x\to -3^{ -}}f(x)$ For $x$ -values smaller than $-3$, the appropriate rule for $f(x)$ is $\ln(-x)+3$. Since $\ln(-x)+3$ is continuous for $x<-3$, any limit in this interval is equal to the expression's value at that limit: $\begin{aligned} &\phantom{=}\lim_{x\to -3^{ -}}f(x) \\\\ &=\lim_{x\to -3^{ -}}[\ln(-x)+3] \gray{\ln(-x)+3\text{ is the rule for }x<-3} \\\\ &=\ln(-(-3))+3 \gray{\ln(-x)+3\text{ is continuous at }x=-3} \\\\ &=\ln(3)+3 \end{aligned}$ Conclusion We found that $\lim_{x\to -3^{ +}}f(x)=\ln(6)$ and $\lim_{x\to -3^{ -}}f(x)=\ln(3)+3$. Since the one-sided limits aren't equal, $\lim_{x\to -3}f(x)$ doesn't exist and $f$ isn't continuous at $x=-3$.